Department of Theoretical Physics,
A. Razmadze Institute of Mathematics,
1 Aleksidze Street, Tbilisi 0193, Georgia

In the present paper geometric aspects of relationship
between non-Noether symmetries and conservation laws in Hamiltonian
systems is discussed. Case of irregular/constrained dynamical systems
on presymplectic and Poisson manifolds is considered.

Non-Noether symmetry; Conservation laws; Constrained dynamics;

70H33, 70H06, 53Z05

Georgian Math. J. 10 (2003) 057-061

Noether's theorem associates conservation laws with particular continuous symmetries of
the Lagrangian. According to the Hojman's theorem [1]-[3]
there exists the definite correspondence between
non-Noether symmetries and conserved quantities. In 1998 M. Lutzky showed that several integrals of
motion might correspond to a single one-parameter group of non-Noether transformations
[4]. In the present paper, the extension of Hojman-Lutzky theorem to singular dynamical systems is considered.

First of all let us recall some basic knowledge of description of the regular dynamical systems
(see, e. g. [5]).
In this case time evolution is governed by Hamilton's equation
$${i}_{{X}_{h}}\omega +dh=0,$$
where $\omega $ is the closed
($d\omega =0$) and non-degenerate
(${i}_{X}\omega =0\Rightarrow X=0$) 2-form,
$h$ is the Hamiltonian and
${i}_{X}\omega $ denotes contraction of
$X$ with $\omega $.
Since $\omega $ is non-degenerate, this gives rise to an isomorphism between the vector
fields and 1-forms given by ${i}_{X}\omega +\alpha =0$.
The vector field is said to be Hamiltonian if it corresponds to exact form
$${i}_{{X}_{f}}\omega +df=0.$$
The Poisson bracket is defined as follows:
$$\{f,g\}={X}_{f}g=-{X}_{g}f={i}_{{X}_{f}}{i}_{{X}_{g}}\omega .$$
By introducing a bivector field $W$ satisfying
$${i}_{X}{i}_{Y}\omega ={i}_{W}{i}_{X}\omega \wedge {i}_{Y}\omega ,$$
Poisson bracket can be rewritten as
$$\{f,g\}={i}_{W}df\wedge dg.$$
It's easy to show that
$${i}_{X}{i}_{Y}{L}_{Z}\omega ={i}_{[Z,W]}{i}_{X}\omega \wedge {i}_{Y}\omega ,$$
where the bracket $[\xb7,\xb7]$ is actually a supercommutator,
for an arbitrary bivector field
$W=\sum _{s}{V}^{s}\wedge {U}^{s}$ we have
$$[X,W]=\sum _{s}[X,{V}^{s}]\wedge {U}^{s}+\sum _{s}{V}^{s}\wedge [X,{U}^{s}]$$
Equation (6) is based on the following useful property of the Lie derivative
$${L}_{X}{i}_{W}\omega ={i}_{[X,W]}\omega +{i}_{W}{L}_{X}\omega .$$
Indeed, for an arbitrary bivector field
$W=\sum _{s}{V}^{s}\wedge {U}^{s}$ we have
$${L}_{X}{i}_{W}\omega ={L}_{X}\sum _{s}{i}_{{V}^{s}\wedge {U}^{s}}\omega ={L}_{X}\sum _{s}{i}_{{U}^{s}}{i}_{{V}^{s}}\omega \phantom{\rule{0ex}{0ex}}=\sum _{s}{i}_{[X,{U}^{s}]}{i}_{{V}^{s}}\omega +\sum _{s}{i}_{{U}^{s}}{i}_{[X,{V}^{s}]}\omega +\sum _{s}{i}_{{U}^{s}}{i}_{{V}^{s}}{L}_{X}\omega ={i}_{[X,W]}\omega +{i}_{W}{L}_{X}\omega $$
where ${L}_{Z}$ denotes the Lie derivative along the vector field $Z$.
According to Liouville's theorem Hamiltonian vector field
preserves $\omega $
$${L}_{{X}_{f}}\omega =0;$$
therefore it commutes with $W$:
$$[{X}_{f},W]=0.$$
In the local coordinates ${z}_{s}$ where
$\omega =\sum _{rs}{\omega}^{rs}d{z}_{r}\wedge {z}_{s}$ bivector field
$W$ has the following form
$W=\sum _{rs}{W}^{rs}\frac{\partial}{\partial {z}_{r}}\wedge \frac{\partial}{\partial {z}_{s}}$ where
${W}^{rs}$ is matrix inverted to ${\omega}^{rs}$.

We can say that a group of transformations
$g\left(z\right)={e}^{z{L}_{E}}$ generated by the vector
field $E$ maps the space of solutions of equation onto itself if
$${i}_{{X}_{h}}{g}_{*}\left(\omega \right)+{g}_{*}\left(dh\right)=0$$
For ${X}_{h}$ satisfying
$${i}_{{X}_{h}}\omega +dh=0$$
Hamilton's equation.
It's easy to show that the vector field $E$ should satisfy
$[E,{X}_{h}]=0$
Indeed,
$${i}_{{X}_{h}}{L}_{E}\omega +d{L}_{E}h={L}_{E}({i}_{{X}_{h}}\omega +dh)=0$$
since $[E,{X}_{h}]=0$.
When $E$ is not Hamiltonian,
the group of transformations $g\left(z\right)={e}^{z{L}_{E}}$ is non-Noether
symmetry (in a sense that it maps solutions onto solutions but does not preserve action).

(Lutzky, 1998) If the vector field $E$ generates non-Noether symmetry,
then the following functions are constant along solutions:
$${I}^{\left(k\right)}={i}_{{W}^{k}}{\omega}_{E}^{k}\text{}k=1...n,$$
where ${W}^{k}$ and ${\omega}_{E}^{k}$ are outer
powers of $W$ and ${L}_{E}\omega $.

We have to prove that ${I}^{\left(k\right)}$ is constant along
the flow generated by the Hamiltonian. In other words, we should find that
${L}_{{X}_{h}}{I}^{\left(k\right)}=0$ is
fulfilled. Let us consider
${L}_{{X}_{h}}{I}^{\left(1\right)}$
$${L}_{{X}_{h}}{I}^{\left(1\right)}={L}_{{X}_{h}}\left({i}_{W}{\omega}_{E}\right)={i}_{[{X}_{h},W]}{\omega}_{E}+{i}_{W}{L}_{{X}_{h}}{\omega}_{E},$$
where according to Liouville's theorem both terms
$[{X}_{h},W]=0$ and
$${i}_{W}{L}_{{X}_{h}}{L}_{E}\omega ={i}_{W}{L}_{E}{L}_{{X}_{h}}\omega =0$$
since $[E,{X}_{h}]=0$ and
${L}_{{X}_{h}}\omega =0$ vanish.
In the same manner one can verify that
${L}_{{X}_{h}}{I}^{\left(k\right)}=0$

Theorem is valid for a larger class of generators $E$ .
Namely, if $[E,{X}_{h}]={X}_{f}$ where ${X}_{f}$ is
an arbitrary Hamiltonian vector field, then ${I}^{\left(k\right)}$ is still conserved. Such a
symmetries map the solutions of the equation
${i}_{{X}_{h}}\omega +dh=0$
on solutions of
$${i}_{{X}_{h}}{g}_{*}\left(\omega \right)+d({g}_{*}h+f)=0$$

Discrete non-Noether symmetries give rise to the conservation of
${I}^{\left(k\right)}={i}_{{W}^{k}}{g}_{*}(\omega {)}^{k}$
where ${g}_{*}\left(\omega \right)$ is transformed $\omega $.

If ${I}^{\left(k\right)}$ is a set of conserved quantities
associated with $E$ and $f$ is any conserved quantity, then the set of functions
$\{{I}^{\left(k\right)},f\}$
(which due to the Poisson theorem are integrals of motion) is associated with
$[{X}_{h},E]$. Namely it is easy to show by taking the Lie
derivative of (15) along vector field $E$ that
$$\{{I}^{\left(k\right)},f\}={i}_{{W}^{k}}{\omega}_{[{X}_{f},E]}^{k}$$
is fulfilled.
As a result conserved quantities associated with Non-Noether symmetries form Lie algebra under
the Poisson bracket.

If generator of symmetry satisfies Yang-Baxter equation
$\left[\right[E[E,W]\left]W\right]=0$ Lutzky's conservation laws are in involution [7]
$\{{Y}^{\left(l\right)},{Y}^{\left(k\right)}\}=0$

The singular Lagrangian (Lagrangian with vanishing Hessian) leads to degenerate 2-form
$\omega $ and we no longer have isomorphism between vector fields and 1-forms.
Since there exists a set of "null vectors" ${u}_{s}$ such that
${i}_{{u}_{s}}\omega =0\text{}s=1,2...n-rank\left(\omega \right),$
every Hamiltonian vector field is
defined up to linear combination of vectors ${u}_{s}$. By identifying ${X}_{f}$
with ${X}_{f}+\sum _{s}{C}_{s}{u}_{s},$ we can introduce equivalence class
${X}_{f}^{\ast}$ (then all ${u}_{s}$ belong to
${0}^{\ast}$ ).
The bivector field $W$ is also far from being unique, but if
${W}_{1}$ and ${W}_{2}$ both satisfy
$${i}_{X}{i}_{Y}\omega ={i}_{{W}_{1,2}}{i}_{X}\omega \wedge {i}_{Y}\omega ,$$
then
$${i}_{({W}_{1}-{W}_{2})}{i}_{X}\omega \wedge {i}_{Y}\omega =0\text{}\forall X,Y$$
is fulfilled. It is possible only when
$${W}_{1}-{W}_{2}=\sum _{s}{v}_{s}\wedge {u}_{s}$$
where ${v}_{s}$ are some vector fields and
${i}_{{u}_{s}}\omega =0$
(in other words when ${W}_{1}-{W}_{2}$ belongs to the class
${0}^{\ast}$)

If the non-Hamiltonian vector field $E$
satisfies $[E,{X}_{h}^{\ast}]={0}^{\ast}$ commutation
relation (generates non-Noether symmetry), then the functions
$${I}^{\left(k\right)}={i}_{{W}^{k}}{\omega}_{E}^{k}\text{}k=1...rank\left(\omega \right)$$
(where ${\omega}_{E}={L}_{E}\omega $) are constant along trajectories.

Let's consider ${I}^{\left(1\right)}$
$${L}_{{X}_{h}^{\ast}}{I}^{\left(1\right)}={L}_{{X}_{h}^{\ast}}\left({i}_{W}{\omega}_{E}\right)={i}_{[{X}_{h}^{\ast},W]}{\omega}_{E}+{i}_{W}{L}_{{X}_{h}^{\ast}}{\omega}_{E}=0$$
The second term vanishes since $[E,{X}_{h}^{\ast}]={0}^{\ast}$ and
${L}_{{X}_{h}^{\ast}}\omega =0$. The first one is
zero as far as $[{X}_{h}^{\ast},{W}^{\ast}]={0}^{\ast}$ and
$[E,{0}^{\ast}]={0}^{\ast}$ are satisfied. So
${I}^{\left(1\right)}$ is conserved.
Similarly one can show that ${L}_{{X}_{h}}{I}^{\left(k\right)}=0$ is
fulfilled.

$W$ is not unique, but ${I}^{\left(k\right)}$ doesn't depend
on choosing representative from the class ${W}^{\ast}$.

Theorem is also valid for generators $E$ satisfying
$[E,{X}_{h}^{\ast}]={X}_{f}^{\ast}$

Hamiltonian description of the relativistic particle leads to the following action
$$A=\int {p}_{0}d{x}_{0}+\sum _{s}{p}_{s}d{x}_{s}$$
where
${p}_{0}=({p}^{2}+{m}^{2}{)}^{1/2}$
with vanishing canonical Hamiltonian and degenerate 2-form defined by
$${p}_{0}\omega =\sum _{s}({p}_{s}d{p}_{s}\wedge d{x}_{0}+{p}_{0}d{p}_{s}\wedge d{x}_{s}).$$
$\omega $ possesses the "null vector field"
${i}_{u}\omega =0$
$$u={p}_{0}\frac{\partial}{\partial {x}_{0}}+\sum _{s}{p}_{s}\frac{\partial}{\partial {x}_{s}}.$$
One can check that the following non- Hamiltonian vector field
$$E={p}_{0}{x}_{0}\frac{\partial}{\partial {x}_{0}}+{p}_{1}{x}_{1}\frac{\partial}{\partial {x}_{1}}+\cdots +{p}_{n}{x}_{n}\frac{\partial}{\partial {x}_{n}}$$
generates non-Noether symmetry. Indeed, $E$ satisfies
$[E,{X}_{h}^{\ast}]={0}^{\ast}$ because of
${X}_{h}^{\ast}={0}^{\ast}$ and $[E,u]=u$.
Corresponding integrals of motion are combinations of momenta:
$${I}^{\left(1\right)}=\sum _{s}{p}_{s}\phantom{\rule{0ex}{0ex}}{I}^{\left(2\right)}=\sum _{r\text{>}s}{p}_{r}{p}_{s}\phantom{\rule{0ex}{0ex}}\cdots \phantom{\rule{0ex}{0ex}}{I}^{\left(n\right)}=\prod _{s}{p}_{s}$$
This example shows that the set of conserved quantities can be obtained from a single
one-parameter group of non-Noether transformations.

Author is grateful to Z. Giunashvili and M. Maziashvili for
constructive discussions and particularly grateful to George Jorjadze for invaluable help.
This work was supported by INTAS (00-00561)
and Scholarship from World Federation of Scientists.

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